Sistemas y Senales: Tarea #3

Cesar A Aceros Moreno

Fecha de asignacion: Feb 10, 2014Fecha de Entrega: Feb 17, 2014

Problemas 1 al 4: Resueltos por el prof. Costas N. Georghiades. Los problemasson sacados de libro: Oppenheim, Willsky and Nawab, Signals & Systems. Problema

5: USTED LO DEBE RESOLVER.

Problema 1 1.27 Linealidad e invarianza en el tiempo de sistemas

continuos

Un sistema puede ser o puede no ser: (a) Sin memoria, (b) lineal, (c) invariable enel tiempo, (d) causal o (e) estable. Para nuestro curso es particularmente importantela linealidad y la invarianza en el tiempo.

Determine, para cada uno de los siguientes sistemas continuos. Cual de estaspropiedades se cumplen y cual no. Presente argumentos que justifiquen sus respuestas.En cada ejemplo, y(t) denota la salida y x(t) la entrada del sistema.

(a) y(t) = x(t− 2)− x(2− t)

(b) y(t) = [cos(3t)]x(t)

(c) y(t) =∫ 2t

−∞x(τ)dτ

(d) y(t) =

{

0, t < 0x(t) + x(t− 2), t ≥ 0

(e) y(t) =

{

0, x(t) < 0x(t) + x(t− 2), x(t) ≥ 0

(f) y(t) = x(t/3)

(g) y(t) = dx(t)dt

. . . . . . . . .

Problema 2 1.28 Linealidad e invarianza en el tiempo de sistemas

discretos

Determine, para cada uno de los siguientes sistemas discretos, cual de las propiedadesenumeradas en el problemas 1.27 se cumple y cual no se cumple. Ofrezca argumentosque justifiquen sus respuestas. En cada ejemp1o y[n] denota la salida y x[n] la entradadel sistema.

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Problem 4 2

(a) y[n] = x[−n]

(b) y[n] = x[n− 2]− 2x[n− 8]

(c) y[n] = nx[n]

(d) y[n] = Ev[x[n− 1]]Ev[x[n]] es la parte par de x[n]

(e) y[n] =

x[n], n ≥ 10, n = 0x[n + 1], n ≤ −1

(f) y[n] =

x[n], n ≥ 10, n = 0x[n], n ≤ −1

(g) y[n] = x[4n+ 1]

. . . . . . . . .

Problema 3 1.31 Utilidad de la propiedad LTI a senales de

entrada complejas

En este problema ilustramos una de las consecuencias mas importantes de las propiedadesde linealidad y de invariancia en el tiempo. Especificamente, una vez que conocemosla respuesta de un sistema lineal o de un sistema lineal invariante en el tiempo (LTI)a una sola entrada o las respuestas a varias entradas, podemos calcular de maneradirecta las respuestas a muchas otras senales de entrada. Gran parte del resto delcurso trata con una amplia explotacion de este hecho para desarrollar resultados ytecnicas para el analisis y sıntesis de los sistemas LTI.

(a) Considere un sistema LTI cuya respuesta a la senal x1(t) en la figura P1.31(a) seala senal y1(t) ilustrada en la figura P1.31(b). Determine y dibuje cuidadosamentela respuesta del sistema a la entrada x2(t) dibujada en la figura P1.31(c).

(b) Determine y dibuje la respuesta del sistema considerado en la parte (a) para laentrada x3(t) mostrada en la figura P1.31(d).

. . . . . . . . .

SISTEMAS Y SENALES SS #

Problem 5 3

Problema4 1.36 Senales periodicas de tiempo continuo y dis-

creto

Sea x(t) la senal continua exponencial compleja:

x(t) = ejw0t

con frecuencia fundamental w0 y periodo fundamental T0 = 2π/w0. Considere lasenal discreta obtenida al tomar muestras de x(t) igualmente espaciadas, esto es,

x[n] = x(nT ) = ejw0nT

(a) Demuestre que si x[n] es periodica si y solo si T/T0 es un numero racional. esdecir, si y solo si algun multiplo del intervalo de muestreo es exactamente iguala un multiplo del periodo x(t).

(b) Suponga que x[n] es periodica, esto es, que

T

T0=

p

q(1)

donde p y q son enteros. Cual es el periodo fundamental y cual la frecuenciafundamental de x[n]? Exprese la frecuencia fundamental como una fraccion dew0T .

(c) Suponiendo nuevamente que T/T0 satisface la ecuacion 1, determine con precisioncuantos periodos de x(t) se necesitan para obtener las muestras que forman unsolo periodo de x[n].

. . . . . . . . .

Problema5 Prueba de Linealidad e Invarianza en el tiempo

Considere los siguientes sistemas y encuentre la respuesta a entradas x1(t) = u(t) yx2(t) = u(t− 1). Determine de sus correspondientes salidas si los sistemas son LIT ono.

(a) y(t) = x(t) ∗ cos(πt)

(b) y(t) = x(t)[u(t)− u(t− 1)]

(c) y(t) = 0.5[x(t) + x(t− 1)]

Dibuje las senales y1(t) y y2(t) para cada caso.

. . . . . . . . .

SISTEMAS Y SENALES SS #

ECEN 314: Signals and Systems

Solutions to HW 3

Problem 1.27

(a) y(t) = x(t− 2) + x(2− t)

Let us check for linearity.

x1(t) → y1(t) = x1(t− 2) + x1(2− t)

x2(t) → y2(t) = x2(t− 2) + x2(2− t)

ax1(t) + bx2(t) = x3(t) → y3(t) = x3(t− 2) + x3(2− t)

= ax1(t− 2) + bx2(t− 2) + ax1(2− t) + bx2(2− t)

= a(x1(t− 2) + x1(2− t)) + b(x2(t− 2) + x2(2− t))

= ay1(t) + by2(t)

Hence linear.

Let us check for time-invariance.

x1(t) → y1(t) = x1(t− 2) + x1(2− t)

x1(t− to) = x2(t) → y2(t) = x2(t− 2) + x2(2− t)

= x1(t− to − 2) + x2(2− t− to)

6= y1(t− to)

Note that y1(t− to) = x1(t− to − 2) + x1(2− t + to). Hence time-variant.

Suppose |x(t)| < B. Then y(t) < B + B = 2B (because |x(t− 2)| < B and |x(2− t)| < B).Hence stable.

Not memoryless as the present output at time t depends on t− 2.

Non-Causal because y(-1)=x(-3)+x(3). So depends on future inputs.

(b) y(t) = [cos(3t)]x(t)

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Let us check for linearity.

x1(t) → y1(t) = [cos(3t)]x1(t)

x2(t) → y2(t) = [cos(3t)]x2(t)

ax1(t) + bx2(t) = x3(t) → y3(t) = [cos(3t)]x3(t)

= [cos(3t)](ax1(t) + bx2(t)

)

= ay1(t) + by2(t)

Hence linear.

Let us check for time-invariance.

x1(t) → y1(t) = [cos(3t)]x1(t)

x1(t− to) = x2(t) → y2(t) = [cos(3t)]x2(t)

= [cos(3t)]x1(t− to)

6= y1(t− to)

Note that y1(t− to) = [cos(3(t− to))]x1(t− to). Hence time-variant.

Stable as |y(t)| < ∞, when |x(t)| < B.

Memoryless as the output at time t depends only on inputs at time t.

Clearly causal.

(c) y(t) =2t∫

−∞x(τ)dτ

Let us check for linearity.

x1(t) → y1(t) =

2t∫

−∞

x1(τ)dτ

x2(t) → y2(t) =

2t∫

−∞

x2(τ)dτ

ax1(t) + bx2(t) = x3(t) → y3(t) =

2t∫

−∞

x3(τ)dτ

2

=

2t∫

−∞

(ax1(τ) + bx2(τ)

)dτ

= a

2t∫

−∞

x1(τ)dτ + b

2t∫

−∞

x2(τ)dτ

= ay1(t) + by2(t)

Hence linear.

Let us check for time-invariance.

x1(t) → y1(t) =

2t∫

−∞

x1(τ)dτ

x1(t− to) = x2(t) → y2(t) =

2t∫

−∞

x2(τ)dτ

=

2t∫

−∞

x1(τ − to)dτ

=

2t−to∫

−∞

x1(τ)dτ

6= y1(t− to)

Note that y1(t− to) =2(t−to)∫−∞

x1(τ)dτ . Hence time-variant.

Suppose x(t) = 1. Then y(1) = ∞. Hence unstable.

Non-causal because, y(1) depends on the value of x(t) at t = 2, as y(1) =2∫

−∞x1(τ)dτ .

Clearly has memory by the above argument.

(d) y(t) =

{0 t < 0

x(t) + x(t− 2) t ≥ 0.

By using the same method as we used for the above parts, it is linear, causal and stable and

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not memoryless. Now let us check for time-invariance.

x1(t) → y1(t) =

{0 t < 0

x1(t) + x1(t− 2) t ≥ 0

x1(t− to) = x2(t) → y2(t) =

{0 t < 0

x2(t) + x2(t− 2) t ≥ 0

=

{0 t < 0

x1(t− to) + x1(t− to − 2) t ≥ 0

6= y1(t− to)

This is because

y1(t− to) =

{0 t < to

x1(t− to) + x2(t− to − 2) t ≥ to

Hence time-variant.

(e) y(t) =

{0 x(t) < 0

x(t) + x(t− 2) x(t) ≥ 0.

By using the same technique as was used for the previous problems, this is time-invariant,not memoryless, stable, causal. Let us check for linearity. Suppose let the input be x1(t) = 1for all t. Then the output y1(t) corresponding to the input x1(t) is

y1(t) = 2. ∀ t

Let us now take the input x2(t) = −x1(t) = −1. If the system is linear, then we shouldget y2(t) = −y1(t) = −2, where y2(t) is the output to the input x2(t). Since x2(t) < 0, theoutput y2(t) = 0 6= −y1(t). Hence not linear.

(f) y(t) = x(t/3) This is linear and stable. It is not memoryless (for example, the out-put at time t = −3 depends on input at t = −1). It is non-causal as well. Let us see whetherit is time-invariant.

x1(t) → y1(t) = x1(t/3)

x1(t− to) = x2(t) → y2(t) = x2(t/3)

= x1

(t

3− to

)

6= y1(t− to)

This is because

y1(t− to) = x1

(t− to3

).

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Hence time-variant.

(g) y(t) = dx(t)dt

This is linear, as well as time-invariant. This is not memoryless as y(t) depends on x(t− δt)

in calculating dx(t)dt

, since dx(t)dt

= limδt→0x(t)−x(t−δt)

δt.

Problem 1.28

(a) y[n] = x[−n]Let us check for linearity.

x1[n] → y1[n] = x1[−n]

x2[n] → y2[n] = x2[−n]

ax1[n] + bx2[n] = x3[n] → y3[n] = x3[−n]

= ax1[−n] + bx2[−n]

= ay1[n] + by2[n]

Hence linear.

Let us check for time-invariance.

x1[n] → y1[n] = x1[−n]

x1[n− no] = x2[n] → y2[n] = x2[−n]

= x1[−n− no]

6= y1[n− no]

Note that y1[n− no] = x1[−n + no]. Hence time-variant.

This is stable as |y[n]| < ∞, if |x[n]| < B.

Non-causal because, y[−1] depends on the value of x[1].

Clearly has memory by the above argument.

(b) y[n] = x[n− 2]− 2x[n− 8]

By using the same technique as used for the above problem, it is linear, time-invariant,causal, stable. This is not memoryless.

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(c) y[n] = nx[n]

x1[n] → y1[n] = nx1[n]

x2[n] → y2[n] = nx2[n]

ax1[n] + bx2[n] = x3[n] → y3[n] = nx3[n]

= n(ax1[n] + bx2[n])

= ay1[n] + by2[n]

Hence linear.

Let us check for time-invariance.

x1[n] → y1[n] = nx1[n]

x1[n− no] = x2[n] → y2[n] = nx2[n]

= nx1[n− no]

6= y1[n− no]

Note that y1[n− no] = (n− no)x1[n− no]. Hence time-variant.

This is not stable because if x[n] = 1 for all n, then y[n] →∞ as n →∞.Memoryless because, y[n] depends only on x[n]. It is also causal.

(d) y[n] = E{x[n− 1]}, where E is the even part.

E{x[n− 1]} =x[n− 1] + x[1− n]

2

This is linear, stable, not memoryless, non-causal. This is time-variant which can be seenby using exactly the same steps as we used for Problem 1.27 (a) with t replaced by n.

(e) y[n] =

x[n] n ≥ 10 n = 0

x[n + 1] n ≤ −1.

This is linear and stable, not memoryless and non-causal. Let us check for time-invariance.

6

x1[n] → y1[n] =

x1[n] n ≥ 10 n = 0

x1[n + 1] n ≤ −1

x1[n− no] = x2[n] → y2[n] =

x2[n] n ≥ 10 n = 0

x2[n + 1] n ≤ −1

=

x1[n− no] n ≥ 10 n = 0

x1[n + 1− no] n ≤ −1

6= y1[n− no]

This is because

y1[n− no] =

x1[n− no] n ≥ no + 10 n = no

x1[n + 1− no] n ≤ no − 1

Hence time-variant.

(f) y[n] =

x[n] n ≥ 10 n = 0

x[n] n ≤ −1.

Following exactly the same steps, it is easy to see from Problem 1.28 (e) that it is linear andtime-variant. It is causal, memoryless and stable.

(g) y[n] = x[4n + 1]

This system is linear as well as stable. Further it is non-causal and memoryless. Let uscheck for time-invariance.

x1[n] → y1[n] = x1[4n + 1]

x1[n− no] = x2[n] → y2[n] = x2[4n + 1]

= x1[4n + 1− no]

6= y1[n− no]

This is becausey1[n− no] = x[4(n− no) + 1] = x1[4n− 4no + 1].

Hence time-variant.

Problem 1.31

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(a) Note that x2(t) = x1(t)−x1(t−2). Therefore using linearity we get y2(t) = y1(t)−y1(t−2).See the figure below.

(b) We see that x3(t) = x1(t + 1) + x1(t). Therefore using linearity we get y3(t) =y1(t + 1) + y1(t). See the figure below.

t0 2 4

y(t)2

−2

2

t

y(t)3

−1 1 2

Problem 1.36

(a) If x[n] is periodic ejωo(n+N)T=ejωonT, where ωo = 2π/To. This implies that

(2π/To)NT = 2πk ⇒ (T/To) = k/N = rational number

(b) If T/To = p/q, then x[n] = ej2πn(p/q). Then the fundamental period is q/gcd(p, q)(from Problem 1.35), and therefore the fundamental frequency is

2π

qgcd(p, q) =

2π

p

p

qgcd(p, q) =

ωoT

pgcd(p, q)

(c) From part (b) above, p/gcd(p, q) periods of x(t) are needed to form a single periodof x[n].

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